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### Divisibility Puzzle

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Date: 01/22/2001 at 23:38:51
From: Li
Subject: Math contest

Dr. Math:

I am a high school student in grade 10 who is in an enriched math
class. Soon I will enter a math contest. Before I ask any specific
questions, could you please give me some suggestions about how to get
myself prepared?

Here is my question: The leftmost digit of an integer of length digits
is 3. In this integer, any two consecutive digits must be divisible by
17 or 23. The 2000th digit may be either "a" or "b" . What is the
value of a+b?

In this question, does consecutive just mean adjacent numbers? If the
number starts with 3, then it will be impossible for any consecutive
number to be divisible by 17 or 23.

I really hope you can point me in a direction to improve my math
solving skills. Are fundamentals more important than techniques?

Yours truly,
A sincere visitor to your Web site
```

```
Date: 01/23/2001 at 16:20:36
From: Doctor Greenie
Subject: Re: Math contest

Hello, Li -

Certainly you can't be successful in mathematics, whether in contests
or in application to a regular job, if you don't have a good, sound
understanding of the fundamentals of mathematics. Although we can
program computers to perform the calculations related to most math
fundamentals, we can't yet program them to think analytically and
devise techniques for solving problems, so in my opinion, techniques
and problem-solving skills are more important than fundamentals.

As for preparation for math contests specifically, my primary
suggestion would be to be confident. Be proud of the math skills you
have, and don't worry about the fact that there are some math skills
you don't have yet. And remember that this is (presumably) only a
friendly competition. If you don't place well in the competition, you
will at least have gained some valuable experience from having
competed.

And one really nice thing about math competitions as I remember them
from my experience at your age is that you have a chance to meet new
friends who share your interest and ability in math.

I'm sure that what they mean is that the 2-digit number formed by any
two adjacent digits in the number is divisible by either 17 or 23.

The 2-digit numbers that are divisible by either 17 or 23 are the
following:

17, 23, 34, 46, 51, 68, 69, 85, and 92

Note that there are two 2-digit numbers in this list with first digit
6 and none with first digit 7; for any other first digit there is only
one entry in the list. (We don't need to worry about first digit 0,
because no 2-digit numbers that are multiples of 17 or 23 have last
digit 0.)  This list of numbers means that the next digit is always
uniquely determined if the current digit is anything other than 0, 6,
or 7. It also means that if the digit 7 appears in the number, then it
is the last digit of the number; and it means that whenever the
current digit is 6, there are two possibilities for the next digit.

Since the first digit is 3, then according to the list of numbers
above, the next digit must be 4, to make the 2-digit number 34.  Then
the third digit must be 6, in order to make the 2-digit number in the
list with first digit 4.

Now, with first digit 6 in the next 2-digit integer, there are two
possibilities for the next digit: 8 or 9. Continuing developing each
of these integers in this manner, we get the following two
possibilities for the integer:

(1) 3468517....
(2) 346923469234692....

The first of these strings leads to a dead end, because there is no
2-digit number divisible by either 17 or 23 with first digit 7.

The second possibility gives us an integer consisting of a sequence of
the 5 repeating digits 34692.

So all the integers that satisfy the conditions of the problem will be
one of two types:

(i) 3469234692....  with these digits repeating to the last digit;

or

(ii) 34692...3468517  with the repeating digits repeating any number
of times but with the number ending in one or more of the
digits "8517" following the final 6

So for most integers that satisfy the conditions of the problem, the
2000th digit is 2; and for a very few such integers (which end shortly
after the 2000th digit) the 2000th digit is 5.

The final answer to the question that was asked is, then, 2 + 5 = 7.

- Doctor Greenie, The Math Forum
http://mathforum.org/dr.math/
```
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